How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr?

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How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr?

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Answer 1 · 2018-05-13T06:25:43

Approx. $50.0*mL$ ....

Explanation:

$"Concentration"="Number of moles"/"Volume of solution"$ ...and so we use this quotient appropriately to find the THIRD value if the other two values are specified.

We want a $200*mL$ volume of $0.500*mol*L^-1$ $NaBr$ ...

$n_"sodium bromide"=200xx10^-3*Lxx0.500*mol*L^-1=0.10*mol.$

And $2.00*mol*L^-1$ $NaBr(aq)$ is available...so....

$(0.10*mol)/(2.00*mol*L^-1)xx1000*mL*L^-1=50.0*mL$ .

Answer 2 · 2018-05-13T06:47:12

$"50.0 mL"$ of the $"2.00 M NaBr"$ solution is needed to make $"200.0 mL"$ of a $"0.500 M NaBr"$ solution.

Explanation:

Use the dilution equation:

$C_1V_1=C_2V_2$ ,

where:

$C_1$ is the initial concentration, $V_1$ is the initial volume, $C_2$ is the final concentration, and $V_2$ is the final volume.

Known

$C_1="2.00 M"$

$C_2="0.500 M"$

$V_2=200.0"mL"$

Unknown

$V_1$

Solution

Rearrange the equation to isolate $V_1$ . Plug in the known values and solve.

$V_1=(C_2V_2)/C_1$

$V_1=(0.500color(red)cancel(color(black)("M"))xx"200.0 mL" )/(0.200color(red)cancel(color(black)("M")))= "50.0 mL"$ (rounded to three significant figures)

$"50.0 mL"$ of the $"2.00 M NaBr"$ solution is needed to make $"200.0 mL"$ of a $"0.500 M NaBr"$ solution.

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How many mL of a 2.00 M NaBr solution are needed to make 200.0 mL of 0.500M NaBr?

2 Answers

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